help-49

[midnight plank]

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[devout burrow]

write the set of positive real numbers between ½ and 100/7 in interval notation form

[storm spindle]

[a,b] is the interval from a to b (including a and b)
(a,b] is the interval from a to b including b but excluding a
[a,b) and (a,b) are defined analogously

Does this help?

[midnight plank]

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[odd sonnet]

i got a question

[odd sonnet]

1. Find the sum up to n terms
   0.6+0.66+0.666+0.6666......

so i did this

and got this

since this sequence is a geometric progression i used the formula Sum=a(r^n-1) /r-1

but this isnt right idk why

[odd sonnet]

is it legal to do this in the first place?

please help guys

<@&286206848099549185>

[coral sage]

Is this the question??

[odd sonnet]

yes

[smoky walrus]

yea that's not right
0.66 ≠ 6*10^-2

[odd sonnet]

yeah I just saw why, my bad

its equal to 0.06

this can be done for 0.6+0.06+0.006... right?

[smoky walrus]

yea

you don't even need to multiply it by 10^n/10^n

[odd sonnet]

yeah, I know how to solve it

I wanted to find another way

[novel lion]

which is?

[odd sonnet]

2/3(0.9+0.99+0.999....)

re writing the sequence

[novel lion]

ah

[odd sonnet]

= 1-0.1+1-(0.1)^2... etc

and you get a gp and a constant sequence

alright thnks

.close

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[odd sonnet]

.reopen

[midnight plank]

✅

[odd sonnet]

what is it/.

[smoky walrus]

there is a way using derivatives

[odd sonnet]

oh really

how

[smoky walrus]

you can write it as
$\sum_{k = 1}^{n} 0.6k10^{-k}$

[grand pond]

Sepdron

[smoky walrus]

doing it as 0.6 + (0.6 + 0.06) + (0.6 + 0.06 + 0.06) + ...

then collect same terms

[odd sonnet]

and then?

[smoky walrus]

then the inside is the same as $- \frac{d}{dx} x^{-k}$ evaluated at x=10

[grand pond]

Sepdron

[smoky walrus]

put the derivative outside of the sum

this is fine because it's a finite sum

then it's just a geometric series

[odd sonnet]

a nice approach

.close

[midnight plank]

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[odd sonnet]

.reopen

?

[midnight plank]

✅

[novel lion]

e?

k*10^(-k-1)

[novel lion]

k*10^(-k)

same?

[midnight plank]

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[manic bison]

tried ax-x^2=0 but that will only give 1 value

nvm i just need someone to explain why ax-x^2+ c = 0 wont have sum of roots as 100 ONLY for a=100

[obtuse bison]

You can factorize ax-x^2 to x*(a-x), which shows you that the two solutions are 0 and a and 100 = 0 + a is trivial to solve 😅

[manic bison]

nah its wrong

i had to take it as npi instead

[obtuse bison]

Yes, but only for n=0, you get real solutions

[manic bison]

no?

nvm ill do it myself ty for trying tho

.close

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[lofty beacon]

Find all prime numbers p and q, such that 5^p+5^q = 0 (mod pq)

My ideas:
p=q => 2p | 2*5^p => p = 5 = q
Now p ≠ q:
p = 5 => 5q | 5^5 + 5^q => q | 625 + 5^(q-1) => q | 626 => q = 2 or 313
Now p≠5, q≠5 and p≠q:
5^q + 5^p = 0 mod p
5^q = -5 mod p
5^(q-1) = -1 mod p
(Symmetrical for mod q):
5^(p-1) = -1 mod q
5^2(q-1) = 1 mod p and 5^2(p-1) = 1 mod q
maybe we can use orders here

[midnight plank]

@lofty beacon Has your question been resolved?

[lofty beacon]

<@&286206848099549185>

[patent peak]

ok so you found the solution for p=q

now suppose that p=5, q≠5

then we require

oh wait nvm you already found this solution

so you have 2 solutions

you just need to consider 2 cases

assume that p<q

if p=2, then 25+5^q=0(mod 2q)

[lofty beacon]

I know all solutions

[patent peak]

the quantity on the left is even, and q≠5, thus 1+5^(q-2)=0(mod q)

[lofty beacon]

But I don’t know how to prove no more

[patent peak]

sorry do you have a third solution?

[lofty beacon]

5,5; 2,5; 2,3; 5,313;

That’s all

Of course we can change (2,3 for 3,2)

[patent peak]

yeah that seems right

so now after you found the solution for p=2

[lofty beacon]

Yes

[patent peak]

we prove that there are no solutions for p is any odd number

[lofty beacon]

Huh?

(3,2)

[patent peak]

like following my statement p<q

without any loss of generality

[lofty beacon]

Ok

[patent peak]

so 5^(q-p)=-1(mod p)

[lofty beacon]

Wait, (313,5)

[lofty beacon]

We can do it without q > p

[patent peak]

ok sorry

[lofty beacon]

Because it be 1/5^n = -1 mod p => -1 = 5^n mod n

[patent peak]

p<q, p≠5, q≠5

[lofty beacon]

p≠q

[lofty beacon]

..

[patent peak]

you can; but i prefer it that way i guess

[lofty beacon]

Ok, we can try

[patent peak]

following from this

[lofty beacon]

Ok

[patent peak]

5^(q-p)=-1(mod q) as well

if you square both, 5^(2q-2p)=1(mod q)

[lofty beacon]

Yep

[patent peak]

5^(2q-2p)=1(mod p)

so ord_5(p)=ord_5(q)

[lofty beacon]

Why?

[patent peak]

wait did i do smth wrong

[lofty beacon]

I think that’s true

But I don’t understand why

[patent peak]

im quite sure they are both equal to 1-v(p-q)

[lofty beacon]

What is v?

[patent peak]

you know the ord() function in modulo arithmetic

[lofty beacon]

No

[patent peak]

hmm

https://www.quora.com/What-are-all-the-primes-p-and-q-that-5-p-5-q-is-divisible-by-pq i found this online i think you can just refer to this

[lofty beacon]

I don’t understand from where he got these

[patent peak]

hmm then i’m not really sure

sorry

[midnight plank]

@lofty beacon Has your question been resolved?

[midnight plank]

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[worthy knoll]

,rotate

[worthy knoll]

just went with normal differentiationg

[lavish venture]

sad

[worthy knoll]

did get it but not in any of the forms given in option

[worthy knoll]

?

[lavish venture]

what did you get

[lavish venture]

i prefer tricks

[worthy knoll]

meh pretty lenthgy

[worthy knoll]

and what might that be?

an hint would suffice... hopefully

[lavish venture]

$\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$

[grand pond]

knief

[worthy knoll]

ahh

[lavish venture]

$F(x,y) : siny - xsin(a+y) = 0$

[grand pond]

knief

[worthy knoll]

why the - ?

[lavish venture]

it’s a theorem

[worthy knoll]

well not taught to us

can u perhaps do it normally?

[lavish venture]

implicit function theorem

[worthy knoll]

this another theorem?

[lavish venture]

yea it’s multivariable calc

but it’s just a shortcut really

[worthy knoll]

im aware of this but no such thing was said under it, ig will be said later

[lavish venture]

you can do normal product rule if you’d like

[worthy knoll]

i did, but not able to convert it into the options

[lavish venture]

cos(y) y’ = x(cos(a+y))y’ + sin(a+y)

then cos(y) y’ - xcos(a+y) y’ = sin(a+y)

[worthy knoll]

shouldnt it be x(cos(a+y))(1 + y')

ahh shit

fuck nvm got it

[lavish venture]

why

[worthy knoll]

well realised my mistake now,, idk y but my brain was processing that a as an x

[lavish venture]

derivative of a + y is y’

[worthy knoll]

will try it again, if i dont get, can i ping u?

[worthy knoll]

yes yes

[lavish venture]

sure

[midnight plank]

@worthy knoll Has your question been resolved?

[midnight plank]

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[worthy knoll]

.reopen

[midnight plank]

✅

[worthy knoll]

yea thanks, got it

.close

[midnight plank]

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[midnight plank]

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[silk violet]

I need help with these 2 questions

[silk violet]

<@&286206848099549185>

[last slate]

!15min

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[prime hornet]

for a), remember your trig ratios and draw a triangle

for b), remember what sec means, and then do the same as in a)

[silk violet]

Oh ok

[prime hornet]

of course, you’ll have to use the unit circle to get all the solutions

[silk violet]

What about this one right here

D and e

[prime hornet]

logarithm rules will probably be helpful

[silk violet]

Yea idk how to do it

Idk where to start

[prime hornet]

,tex .log rules

[grand pond]

higher!

[prime hornet]

when you take the log of both sides, you’re going to be applying the first three rules here

then it’s just algebra to solve for x

[silk violet]

I'll get back to that

I'm still stuck on question 8 a

This makes 0 sense

[silk violet]

.

Yea I'm so lost

[silk violet]

???

Can someone actually help?

[fallow scarab]

use power rule log property

but product first

[silk violet]

How

[fallow scarab]

a * b^x = 50 * (2/3)^x

a = 50, b = 2/3

if you set c = b^x, then log(ac) = log(a) + log(c)

that's what i mean product first

[silk violet]

Ohh ok

So leave the 11

What's c?

Is it 11?

[silk violet]

Idk where ur getting those letters from

Where's c

And x

[fallow scarab]

i'm just abstract away numbers from your problem so you can identify the patterns so you can use this table

[fallow scarab]

and i'm using Pemdas to identify different orders of operations that's happening

https://www.mathsisfun.com/operation-order-pemdas.html

[silk violet]

Oh ok

[midnight plank]

@silk violet Has your question been resolved?

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[last slate]

is this related to math?

it's about Pokémon

[cursive vine]

.close

[midnight plank]

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[clever niche]

What i did ^

Question:

Not sure where i went wrong

[midnight plank]

@clever niche Has your question been resolved?

[wintry bane]

You forgot to multiply the 2 in numerator with 4pi

4pi = 12.566 but your numerator was 2*4pi

[clever niche]

oh so what i did was 2pi squared /0.4 pi squared

but for numerator i should have done 4 pi squared?

[wintry bane]

No no you did everything right

Until the last step

[clever niche]

OH

i didnt x2

[wintry bane]

When you calculated the values, you put 12.566 in numerator but it's wrong

yess

[clever niche]

ohh

so i ignored it like here

[wintry bane]

Yea

[clever niche]

did (pi x 2^2) x2

[wintry bane]

Yupp

[clever niche]

the x2 i didnt do i left ot

omg i feel a bit stupid lol

thank you

i was proper stressing

[wintry bane]

haha np

[clever niche]

.close

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[full obsidian]

In this picture, A belongs to alpha, B belongs to betta, C belongs to straight line L. (A€a, B€b, C€l) What is the error in this picture? Give an explanation

Pls help me

[midnight plank]

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@full obsidian Has your question been resolved?

[midnight plank]

@full obsidian Has your question been resolved?

[midnight plank]

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[thorny delta]

How do i integrate this with the tan inverse thing ... quick response would be much appreciated as i have exam in a few hours : -) please help

[merry pewter]

factor out the 4 in the fractional part and do a u-sub

[thorny delta]

like this ?

[merry pewter]

yeah except you need another 2 because of chain rule

unless that it a u

in the third line

[thorny delta]

now ok?

[merry pewter]

yes

[thorny delta]

thank you!!!!

.close

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[upbeat plinth]

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[tribal tartan]

well y=sqrt(2) is a horizontal line

a line that would be perpendicular to a horizontal line would be a vertical line

that still passes through the same exact coordinate

[upbeat plinth]

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[tribal tartan]

yep thats it

[upbeat plinth]

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[midnight plank]

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[upbeat plinth]

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[blissful totem]

can you avoid closing channel and reopening for the same question jsut to get top of channels

[upbeat plinth]

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[torn flint]

Derivative?

[upbeat plinth]

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[torn flint]

Well what is the derivative

[upbeat plinth]

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[torn flint]

No, what does it mean

[upbeat plinth]

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[torn flint]

Not kind of

So you’re looking for when the slope of the tangent line is 6

You’re looking for when the derivative is 6

[upbeat plinth]

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[fathom knoll]

It's maybe important to just add explicitly (to be safe) that the slope of the tangent line is precisely the derivative of the function at that point

[torn flint]

At what values of x will 6x^2 be 6?

[upbeat plinth]

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[torn flint]

And?

[upbeat plinth]

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[fickle oriole]

Lock in kenzo

[upbeat plinth]

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[torn flint]

It’s asking for coordinates. You have the x values, how do you find the y values?

[upbeat plinth]

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[torn flint]

Try it

[upbeat plinth]

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[torn flint]

And?

[upbeat plinth]

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[torn flint]

Therefore?

[upbeat plinth]

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[torn flint]

Right

Good job

[upbeat plinth]

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[midnight plank]

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[midnight plank]

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[hot stirrup]

Im so confused on how to do part c

[hot stirrup]

Even though it looks like a bernoulli distribution isnt it a binomial?

<@&286206848099549185>

.close

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[timid tangle]

Hello

[timid tangle]

Is this the correct answer?

Can you explain c and b?

[sharp coral]

there is one which is false

[timid tangle]

c?

[sharp coral]

if a claim is false you should be able to find a counterexample

consistent means has at least 1 solution, yes

[timid tangle]

inconsistent can also mean infinite solutions, no?

[sharp coral]

no, inconsistent means no solutions

is it? can you find a counterexample?

is it?

[timid tangle]

Ohhh ic

Thank you so much guys

.close

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[fierce canyon]

AC=AB+BD

That’s the only thing given

Find x

C is cut out btw

[midnight plank]

@fierce canyon Has your question been resolved?

[fierce canyon]

<@&286206848099549185>

ples

[wet bridge]

on it

[fierce canyon]

thx

[viral dagger]

i swear ive seen this question before

[fierce canyon]

idk my teacher just randomly gives questions out

[viral dagger]

oh nvm its a diffrent one but very simmilar

[wet bridge]

extend CB to a point P such that PB=BA
PBA is isosceles so <CPA= 40
that makes the triangle PDA isosceles
therefore PA=PD=a+b and since AC=a+b, that makes the triangle PAC isosceles as well
so x=40

my drawing is not to scale

[wet bridge]

.

[fierce canyon]

oh i get it tysm but theres one more question i was wondering if u could help me with

[wet bridge]

no promises but feel free to send

[fierce canyon]

a and b are consecutive even numbers c=b/a-a/b give the value of c in terms of a and explain that this will always be even/even i put a as 2x and b as 2x+2 but when i simplified it i got (2x+1)/(x^2+x) but 2x+1 is odd. When i did a=x and b=x+2 i got (4x+4)/(x^2+2x) which is even/even is there a reason for this

[wet bridge]

yeah im not good at algebra

[fierce canyon]

o

[fierce canyon]

@viral dagger do u think u could help with this

[viral dagger]

the first one is just a more simplified version of the second one (assuming you substitute something like x=2k so it is for all integers k)

the first one works for all integers, the second one only even x

[fierce canyon]

wait wdym by that?

cuxz the results are kinda different

[viral dagger]

well if its x and x+2 as the consecutive even integers, that means x has to be even aswell right?

uhh

[fierce canyon]

but if 2x is even then 2x+1 has to be odd and that isnt even/even

[viral dagger]

6/8 can be simplified to 3/4 and 3 is odd so?

[fierce canyon]

so how do we classify it as even/even cause that means that like all fractions are even/even right

im kinda confused

[viral dagger]

your asking why the first one is odd/even and the second os even/even right?

[fierce canyon]

Yeah

And like if you can just unsimplify it

For even/even

idk I’m lost

[viral dagger]

what do you mean unsimplify it

tldr it just doesent matter, say the first one was 3/4 and the second one was 6/8, they equal eachother so it doesent matter

the second one is just unsimplified

[fierce canyon]

So does the first one not work then even if it’s the same cause it’s even/even

[viral dagger]

both works

but use the first one

[fierce canyon]

But isn’t it not technically even/even

[viral dagger]

??

[fierce canyon]

cause 3/4 is odd/even

[viral dagger]

its this

eugh how do i explain it

basically the second one accounts for odd and even integers, so it cant be simplified further

try plugging in x=2k into the second equation so you only get even integers x, after simplifying you will see that its judt the first equation

[fierce canyon]

so the first one works for any number but the second one is only even? i dont rly understand

[viral dagger]

yeah

basically, you wrote a=x and b=x+2, if you see here if x is odd then a is also odd, and if x is even then a is also even, since a has to be even then x has to be even

[fierce canyon]

so if a=2x then x can be odd or even

[viral dagger]

yes, but a is always even

[fierce canyon]

wait so if x is even in the first one then its even but if x is odd in the first one then its odd

[viral dagger]

yes

[fierce canyon]

but for the second one it has to be both even so its both even

[viral dagger]

yes

[fierce canyon]

im kinda just processing it

idkk its rly confusing

[fierce canyon]

so how would i answer it cause it says always even/even but im answering with an odd/even

[viral dagger]

sub 2x=a in the first eq, thats it

[fierce canyon]

so a+1/(1/2(a^2)+1/2a)

[fierce canyon]

wait ik u kinda said it a lot but i still dont really get it cause a+1 would be odd and not ALWAYS even 😭 srryy

[viral dagger]

multiply top and bottom frac by 2

[viral dagger]

it does not matter weather a+1 is odd or even

[fierce canyon]

why not?

[fierce canyon]

but then you can just make any fraction even right

[viral dagger]

literally im guessing your confused on why even/even+even/even can get odd/even, its literally just plain simplifying

[viral dagger]

yes

[fierce canyon]

so youre saying you can get even/even from an odd/even?

[viral dagger]

yes

just multiply top abd bittom by 2

[fierce canyon]

and thats why its always even/even?

[fierce canyon]

im mainly confused cause it says always even/even but its not when simplified

[viral dagger]

any even number can be written as 2k for all integers k, so 2k/2l for integers k and l can be simplified to k/l

here k just happens to be odd

[fierce canyon]

k could be either right

not just odd for this

[fierce canyon]

can u elaborate a bit with like the example in the question

[viral dagger]

alr how about this

lets say the number is even/even, then they must have a common factor of 2, so if its odd/even then its diffrent from even/even, if its even/odd then its diffrent from even/even, if its still even/even then they share a common factor of 2 again

if you repeat this over and over again its only either even/odd or odd/even

[fierce canyon]

wait so then wouldnt every fraction be even/even

[viral dagger]

you can change every fraction to be even/even

[fierce canyon]

so how would the question even work if its always true if you mutliply 2 so we dont need algebra

[viral dagger]

what does that have to do with anything??

the fact that weather its even/even is not important to this question at all 😭

[fierce canyon]

isnt the question talking about even/even?

[viral dagger]

yes but the fact that one is odd/even and one is even/even does not matter

[fierce canyon]

wait wdym?

mb my brain is not working tdy

[viral dagger]

this, hs first being odd/even and the second being even/even doesent matter

[fierce canyon]

theyre both even/even technically right

[viral dagger]

yes

[fierce canyon]

wait so how do we answer this question?

[viral dagger]

multiply by 2 on top and bottom to get (2(a+1))/(a(a+1))
and just simplify to get 2/a

[fierce canyon]

wait why do we multiply this by 2?

[fierce canyon]

.

mb forgot to reply

[viral dagger]

to get rid of fractions

[fierce canyon]

so just to help simplify it

[midnight plank]

@fierce canyon Has your question been resolved?

[midnight plank]

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[tacit kelp]

Hi

[tacit kelp]

I need help please especially with the third and fourth photo

[last slate]

i thaught u gonna send new photo

sorry

[tacit kelp]

@last slate if ur busy its fine,

I dont want to disturb you 🙂

But if ur not busy i would love some help 🙂

I tried solving it my self is it correct?

I’ve got to go, but thanks for ur help,

.close

[midnight plank]

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[last slate]

hello

[grand pond]

Svaha

[last slate]

from LHS to RHS

[midnight plank]

@last slate Has your question been resolved?

[blissful totem]

the derivation of compound angle formulas is best done geometrically

theres multiple variation of them

i think its best u just google or search them up on youtube

video format will do a much better than job text in showing these proofs

[midnight plank]

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[twilit field]

I'm trying to prove that $\gcd(a,b)$ is unique

[grand pond]

Veni, vidi, perii is $\R - \Q$

[twilit field]

So my idea is let the gcd be, if possible, both $k_1$ and $k_2$. We then have $ k_1\geq k_2$ and $k_2 \geq k_1$ from the definition of $(\gcd(a,b)$, from which it trivially follows that $k_1=k_2$

[grand pond]

Veni, vidi, perii is $\R - \Q$

[twilit field]

Is this fine?

[runic hamlet]

yes

[twilit field]

Thanks!

[visual tiger]

it's like saying the maximum of a set is unique

[twilit field]

Yeah, I know

[last slate]

i didn't find them

.reopen

huh

awesome

[twilit field]

Prove that $lcm(a,b) = \frac{ab}{\gcd(a,b))}$
\
We know that $ a\mid \frac{ab}{\gcd(a,b)}$ and $ b \mid \frac{ab}{\gcd(a,b)}$
\
We now assume $a \mid e$ and $b \mid e$., where $e$ is a common multiple of $a$ and $b$
\
we wish to prove that $e \geq \frac {ab}{\gcd(a,b)}$
It then follows that $e^2=abk_1k_2$
\
From this it follows that $e^2\geq e \geq ab$
\
It also follows that $e \geq \gcd(a,b)$
\
so $\frac{e^2}{\gcd(a,b)} \geq e \geq \frac{e}{\gcd(a,b)} \geq \frac {ab}{\gcd(a,b)}$
\

[grand pond]

Veni, vidi, perii is $\R - \Q$

[twilit field]

Prove that $lcm(a,b) = \frac{ab}{\gcd(a,b))}$
\
We know that $ a\mid \frac{ab}{\gcd(a,b)}$ and $ b \mid \frac{ab}{\gcd(a,b)}$
\
We now assume $a \mid e$ and $b \mid e$., where $e$ is a common multiple of $a$ and $b$
\
we wish to prove that $e \geq \frac {ab}{\gcd(a,b)}$
It then follows that $e^2=abk_1k_2$
\
From this it follows that $e^2>ab$
\
It also follows that $e \geq \gcd(a,b)$
\
$ab \geq gcd(a,b)$ also follows
\
so $e^2 \geq e \geq ab \geq gcd(a,b)$
\

[grand pond]

Veni, vidi, perii is $\R - \Q$

[twilit field]

Prove that $lcm(a,b) = \frac{ab}{\gcd(a,b))}$
\
We know that $ a\mid \frac{ab}{\gcd(a,b)}$ and $ b \mid \frac{ab}{\gcd(a,b)}$
\
We now assume $a \mid e$ and $b \mid e$., where $e$ is a common multiple of $a$ and $b$
\
we wish to prove that $e \geq \frac {ab}{\gcd(a,b)}$
It then follows that $e^2=abk_1k_2$
\
From this it follows that $e^2>ab$
\
It also follows that $e \geq \gcd(a,b)$
\
$ab \geq gcd(a,b)$ also follows
\
so $e^2 \geq ab \geq gcd(a,b)$
\
It's trivially true that $ab \geq \frac{ab}{\gcd(a,b)}$
\|
There are two possible cases.
\
$e^2 \geq e \geq ab \geq \gcd(a,b)$
\
Or
\
$e^ \geq ab \geq e \geq gcd(a,b)$
so

[grand pond]

Veni, vidi, perii is $\R - \Q$

[twilit field]

Prove that $lcm(a,b) = \frac{ab}{\gcd(a,b))}$
\
We know that $ a\mid \frac{ab}{\gcd(a,b)}$ and $ b \mid \frac{ab}{\gcd(a,b)}$
\
We now assume $a \mid e$ and $b \mid e$., where $e$ is a common multiple of $a$ and $b$
\
we wish to prove that $e \geq \frac {ab}{\gcd(a,b)}$
It then follows that $e^2=abk_1k_2$
\
From this it follows that $e^2>ab$
\
It also follows that $e \geq \gcd(a,b)$
\
$ab \geq gcd(a,b)$ also follows
\
so $e^2 \geq ab \geq gcd(a,b)$
\
It's trivially true that $ab \geq \frac{ab}{\gcd(a,b)}$
\
There are two possible cases.
\
$e^2 \geq e \geq ab \geq \gcd(a,b)$
\
Or
\
$e^ 2 \geq ab \geq e \geq gcd(a,b)$
\
In the first cases
\
$\frac{e}{\gcd(a,b)} \geq \frac{ab}{\gcd(a,b)} \geq 1$
\
In the second case, $\frac{e^2}{\gcd(a,b)} \geq \frac{ab}{\gcd(a,b)} \geq \frac{e}{\gcd(a,b)}\geq 1$

[grand pond]

Veni, vidi, perii is $\R - \Q$

[twilit field]

Prove that $lcm(a,b) = \frac{ab}{\gcd(a,b))}$
\
We know that $ a\mid \frac{ab}{\gcd(a,b)}$ and $ b \mid \frac{ab}{\gcd(a,b)}$
\
We now assume $a \mid e$ and $b \mid e$., where $e$ is a common multiple of $a$ and $b$
\
we wish to prove that $e \geq \frac {ab}{\gcd(a,b)}$
It then follows that $e^2=abk_1k_2$
\
From this it follows that $e^2>ab$
\
It also follows that $e \geq \gcd(a,b)$
\
$ab \geq gcd(a,b)$ also follows
\
so $e^2 \geq ab \geq gcd(a,b)$
\
It's trivially true that $ab \geq \frac{ab}{\gcd(a,b)}$
\
There are two possible cases.
\
$e^2 \geq e \geq ab \geq \gcd(a,b)$
\
Or
\
$e^ 2 \geq ab \geq e \geq gcd(a,b)$
\
In the first cases
\
$\frac{e}{\gcd(a,b)} \geq \frac{ab}{\gcd(a,b)} \geq 1$
\
In the second case,
\

[grand pond]

Veni, vidi, perii is $\R - \Q$

[twilit field]

<@&286206848099549185> can I have help with case 2

The best I can do is $\frac{e^2}{ab} \geq 1 \geq \frac{e}{ab} \geq \frac{gcd(a,b)}{ab}$

[grand pond]

Veni, vidi, perii is $\R - \Q$

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

@twilit field Has your question been resolved?

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[dense arch]

can anyone help me with the a? i need to write the proof of the limit of a sequence as well

[midnight plank]

@dense arch Has your question been resolved?

[midnight plank]

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[devout cove]

is it possible to divide this expression?

[devout cove]

the original problem is

I just expanded those two as a^3 - b^3

[umbral scroll]

hmm

[devout cove]

pretty sure I made a mistake from the beginning

you can't just multiply them

how should I start this?

[umbral scroll]

so you are looking to develop this ?

[devout cove]

yes, but I think using a^3 - b^3 is a mistake

[umbral scroll]

uhm

[devout cove]

this now equals to 2x - 1

which does not correspond to the original expression

[umbral scroll]

yes it looks kinda hard to calculate

why not

[devout cove]

2x - 1 != cbrt(2x) - 1

[umbral scroll]

uh no

[devout cove]

I should have multiplied on both sides

[umbral scroll]

i think you gotta use a^3 -b^3 actually

i dont see another way

[devout cove]

yes but

I multiplied it wrong

I have to multiply this by the numerator and the denominator

if this was 2x - 1, it would be correct

[umbral scroll]

do the denominator and the numerator ^3

uh no

[devout cove]

what do you mean?

cube them all?

[umbral scroll]

nah it wont work

[devout cove]

I don't think u can do that

[umbral scroll]

it wont remove the cubic roots

you could try separating the two elements of the numerator

[twin forge]

How did you do a^3 - b^3 here?

[devout cove]

this is the way to go

[devout cove]

cbrt(4x^2 - 1)

you can multiply by (cbrt(16x^4) + cbrt(4x^2) + 1)

that'll get rid of the cube

yea it'll be 4x^2 - 1 on the numerator

but I think I should have gotten rid of the cube root

on the denominator

not the numerator

[twin forge]

Do the same for cbrt(2x) -1

[devout cove]

what's the point?

it's being multiplied by cube roots anyway

[twin forge]

You can cancel 2x-1

4x^2 - 1 = (2x-1)(2x+1)

[devout cove]

that's true

but

what do I do about this

it's being multiplied by the factor of cbrt(4x^2) - 1

not only that, if I factor cbrt(2x ) - 1, I am gonna get cubes on the numerator anyway

I think I should have factored the denominator first

[twin forge]

Oh

[twin forge]

Take the LCM here

Much easier that way

[devout cove]

lcm?

lcm of what though?

[twin forge]

$\frac{\sqrt[3]{4x^2} - 1}{\sqrt[3]{2x} - 1} - 1$

[grand pond]

Adarsh

[twin forge]

$= \frac{\sqrt[3]{4x^2} - \sqrt[3]{2x} }{\sqrt[3]{2x} - 1}

[devout cove]

$= \frac{\sqrt[3]{4x^2} - \sqrt[3]{2x} }{\sqrt[3]{2x} - 1}$

[grand pond]

asm

[twin forge]

Take $\sqrt[3]{2x}$ common

[grand pond]

Adarsh

[devout cove]

wait did you get common denominators with that -1?

[twin forge]

Yea

[devout cove]

or where'd the -1 go

wouldn't it be -1 - cbrt(2x)

oh

I see

yea let me try that

$\frac{cbrt(2x)(cbrt(4x) - 1)}{cbrt(2x) - 1}$

[grand pond]

asm

[devout cove]

is it possible to cancel now?

[twin forge]

$= \sqrt[3]{2x} * \frac{\sqrt[3]{2x} - 1 }{\sqrt[3]{2x} - 1}$

[grand pond]

Adarsh

[twin forge]

Wouldnt it be this?

As $\sqrt[3]{2x} * \sqrt[3]{2x} = \sqrt[3]{4x^2}$

[grand pond]

Adarsh

[devout cove]

oh

I meant

cbrt(2x)

yea you can just cancel now

$(sqrt[3]{2x})^6$

[grand pond]

asm

[devout cove]

I mean

that'll just be the answer right

I really don't wanna raise 2 to the 6th power 🤣

i'll multiply it by 1/x^2 and thats it

[twin forge]

$(\sqrt[3]{2x})^6 = (2x)^2$

[grand pond]

Adarsh

[devout cove]

oh I see

1/3 * 6 = 2

alright, thank yuo so much 👍

.close

[midnight plank]

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[topaz portal]

The first of four consecutive odd integers is x. The product of 45 and the third integer is at least 84 less than the sum of the first and fourth integers. What is the greatest possible value of x?

[umbral scroll]

can they be equal

[pale nacelle]

can some help me with radical in maths im in the 8 grde but pls in call

[umbral scroll]

ig you have two inequations

[topaz portal]

I set it like 45(x+4)>=(x+(x+6))-84

Idk how to make it 2

[umbral scroll]

well

what i see is

x< x+2 +x+4 + x+6

and

45(x+4) > x + x+6 - 84

yes

[topaz portal]

What would x be then

[umbral scroll]

well you can solve it

it is a simple inequation

you wont find the val of x but x> smth

or x>smth

[midnight plank]

@topaz portal Has your question been resolved?

[midnight plank]

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[golden turret]

oops wrong picture one sec

Wouldn't h be congruent to g under mod p?

and thus they are the same root?

[runic hamlet]

yes but mod p^2 they are different

and so they^(p-1) can have different results mod p^2

[golden turret]

so just to check the theorem is different from saying that there exists primitive root g in {0,1,...,p-1} st g^{p-1} is not congruent to 1 mod p^2

[runic hamlet]

if by {0,1,...,p-1} you mean the actual numbers, then yes

[midnight plank]

@golden turret Has your question been resolved?

[midnight plank]

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[last slate]

uhh

[last slate]

open channel?

oh shit

[umbral scroll]

yes tell me

[last slate]

what the hell

gonna copy and paste my message rq one second sorry

can you help with my math re-entry exam so i don’t have restart junior year at 17 bc ive spent 8 months in a group home and have forgotten how to do algebra. i just need to relearn it and im not good at watching videos bc i need to ask questions.

i suck at numbers bc i’m a history nut who’s only good with dates

[umbral scroll]

any specific question ?

[last slate]

lemme think of an example

[umbral scroll]

honesly$

[last slate]

it’s probably gonna sound stupid

[umbral scroll]

so your issue is with calculus right ?

[last slate]

one moment while i pull up the program

[last slate]

algebra

[umbral scroll]

ok

[last slate]

and it looks so simple too

[umbral scroll]

you know

[last slate]

?

[umbral scroll]

you could ask your teacher for a book to train your calculus and algebra

[last slate]

it’s online school

[umbral scroll]

like to solve equations

[last slate]

programs called acellus

but anyways

an example would be uhh

4 - 2x
——— =4
3

how the fuck do i solve for x

[mellow sand]

we wanna get the x alone

[last slate]

hey there x… want some candy?

[mellow sand]

no

like on one side of the equation

[last slate]

ohhh

sooo

4 - 2
——— =4x
3

?

[mellow sand]

multiple both sides by 3

[gaunt ibex]

follow order of operations

[mellow sand]

we don’t like fractions

[last slate]

so both 4s multiplied by 3?

[last slate]

pemdas

uhhhh

4-2x = 12?

or 12-2x=12

i’ve been abandoned by my saviors

[mellow sand]

this

[last slate]

ok

now what

[mellow sand]

subtract 4 from both sides

[last slate]

2x = 8

wait no

don’t i keep the -

?

if so then -2x = 8

[mellow sand]

yep

[last slate]

but now i’m lost

[mellow sand]

then divide by -2

[last slate]

-2x = -4x?

wait no

sorry

-2x = -4

[mellow sand]

no

-2x = 8 right

if we divide by -2 we get

x = -4

[last slate]

ohhh shit

[mellow sand]

understand?

[last slate]

ok i think i got that type down then

help me figure out the other type?

also fellow climber

nice

[last slate]

i’ll just put it down anyways

8(x-8)=16

i think that would work

i’m just using numbers that fit into each other

[mellow sand]

is this the problem?

[last slate]

yes

[mellow sand]

okay

do you know what to do first?

[last slate]

uhhh

you said we don’t like parentheses

so those

wait no the x is in there

my bad

[mellow sand]

we could either divide by 8 or distribute the 8

i would recommend dividing by 8

what do you get then?

[last slate]

the 16 right?

then 2

8(x-8)=2

[mellow sand]

you divided by 8 so what happens to the 8 on the left?

[last slate]

gone

vanished

x-8=2

[mellow sand]

okay so what is the actual equation?

yep

now if we wanna get x alone what must we do

[last slate]

uhhh

8-2?

errr

2-8

but the 8s a negative

[mellow sand]

we need to add 8 to both sides

[last slate]

so unless i’m stupid then it’s positive 6!

fuck

[mellow sand]

cause opposite of - is +

so to get rid of a - you need to add

[last slate]

x-16=10

?

[mellow sand]

nope

you add 8 to both sides

[last slate]

i did :(

[mellow sand]

you added 8 to the right and subtracted 8 from the left

[last slate]

but the left was 8

[mellow sand]

x - 8 = 2

[last slate]

8+8 is 16

[mellow sand]

-8 + 8 = 0

-*

weird discord

[last slate]

x-0=10

[mellow sand]

or x = 10

[last slate]

uhhh

wait what

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[golden turret]

Im not sure how we know that we can reduce x^4 congrunet to 3^4 mod 17 to x^2 congruent to +-9 mod 17

[carmine sigil]

Well, the simple answer is the integers mod 17 comprises a finite field.

But that might be unsatisfying

as you likely haven't been introduced to that concept yet.

[golden turret]

hmm yes im not sure what that means yet. but generally is the statement x^4 congruent to q^4 under mod p^4 implies x^2 is congruent to q^2 under mod p^2 where p and q are prime true?

[carmine sigil]

I'm pretty sure that q does not need to be prime

and p only needs to be the power of a prime.

well

no p needs to be prime

you can make a field with a power of a prime, but it requires doing something weird with multiplication

[golden turret]

the same theorem is applied here to obtain solutions to x^2 \equiv 9 mod 17 is +-3 right?

[carmine sigil]

so we have 3^4 = x^4 mod 17 So this is just an algebra problem, so we would expect 4 solutions.

x^4 = 81 mod 17 implies x^2 = 9 mod 17 or x^2 = -9 mod 17 = 8 mod 17

we can compute both of these values to find that 8 * 8 = 64 mod 17 = 13 mod 17, and 9 * 9 = 81 mod 17 = 13 mod 17 as expected.

Then for each of these we need to find the root mod 17, 9 is easy as +/- 3, but 8 is tricker.

we need some k such that 8 + 17k is a square number, and it turns out that k = 1 does the trick

so 8 + 17 = 5, so we have 5 mod 17 and -5 = 12 mod 17

so the roots are, 3, 5, 12, and 14

does the above make sense?

@golden turret

[golden turret]

ah

okay

i think i got part of it, just need some time

thanks for the help

.close

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[velvet coyote]

can someone help me out pls

[wet pollen]

You didn't dive by v^2

[velvet coyote]

v^2?

[wind cliff]

U could also do without the quotient rule and just use the product

[wet pollen]

The denominator squared.

[velvet coyote]

OMG

i completely forgot that part omg

ok wait let me do that again gimme a sec

[spring latch]

pro tip: quotient rule is easily forgettable and easy to make mistakes on, so just use product rule with 1/g(x)

[velvet coyote]

he wants us to learn the quotient rule for the class so i'm js gonna do it tjay way thanks though

[velvet coyote]

do i have to expand the denominator like i did in the pic ? in some examples in class we didn't do that so idk

[barren echo]

That doesn't matter I think
U need to find the equation of the tangent anyway

[velvet coyote]

how do i do that after finding f'(x)?

[wet pollen]

Use the x-component of the given point (3,4) and plug that into f'(x). That will give you the derivative/slope at (3,4).

[velvet coyote]

ahh okay

what would i get if i plugged in the y ?

oh

stupid question LMAO

[velvet coyote]

lemme do it

[velvet coyote]

-7/3?

[wet pollen]

,wolf derivative f(x)=(5x-3)/(3x-6)

[grand pond]

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[wet pollen]

,wolf derivative f(x) = (5x-3)/(3x-6) at x=3

[grand pond]

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Upgrade to WolframAlpha Pro!

Requested by kookiemon

[wet pollen]

Looks good.

[velvet coyote]

how do i set up the equation ?

[wet pollen]

Point-Slope formula

y - y_1 = m*(x - x_1)

[velvet coyote]

so y - 4 = -7/3 ( x - 3)?

[wet pollen]

Yes.

[velvet coyote]

okay thank you so much !!

[wet pollen]

yw

[velvet coyote]

.close

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[quaint field]

could I use sin in this formula instead of dot product with the normal vector?

it makes sense to me because sin maximizes the value at perpendicular
sin(90) = 1

specifically sin of the angle between the curve and the vector field

I feel like I am getting the same values both ways

[midnight plank]

@quaint field Has your question been resolved?

[midnight plank]

@quaint field Has your question been resolved?

[midnight plank]

@quaint field Has your question been resolved?

[midnight plank]

@quaint field Has your question been resolved?

[midnight plank]

@quaint field Has your question been resolved?

[dim moss]

I mean... given that cos(x) = sin(pi/2 - x), it checks out...

maybe this doesn't generalize easily to 3 dimensions or something?

if phi is the easier one to find, then I don't see any reason why not :)

[midnight plank]

@quaint field Has your question been resolved?

[quaint field]

yeah that makes sense

I guess they use cosine too because that can be indirectly calculated using dot product

now my question is how would I find the normal vector?

[queen herald]

If $(x(t), y(t))$ is your contour, then the normal vector is equal to $\frac{1}{\sqrt{\dot{x}^2 + \dot{y}^2}} (-\dot{y}, \dot{x})$

[grand pond]

EQUENOS

[quaint field]

oh, so it’s just flipping the signs?

I thought you had to take the second derivative or something

because wouldn’t acceleration be normal to velocity

[queen herald]

In circular motion yes, but not all curves are circles

For example consider $(x(t), y(t)) = (t^2, 0)$

[grand pond]

EQUENOS

[queen herald]

The acceleration and velocity end up being collinear

[quaint field]

hmm yeah

it only is perpendicular is velocity is constant while direction changes

like in circular motion

so I guess you can directly flip the signs to get the normal vector

@queen herald is that all I need to solve this problem?

also is there an online tool I can use to check my answer

[queen herald]

yes

[queen herald]

Most likely wolfram alpha will do

[quaint field]

is there no vector field graphing tool?

where I can define a vector field and a curve

and it'll tell me the flux

through numerical calculation or something

[queen herald]

Once again, wolfram alpha

[quaint field]

oh

[quaint field]

wait, aren't there two possible normal vectors (with opposing directions)? how would I know which one to use

I guess the one pointing outwards from the circle

but how would I know that

[quaint field]

cool

[queen herald]

If you (x(t), y(t)) runs through the circle clockwise, then (-y', x') will point outwards

[quaint field]

will the direction affect the answer?

or can I use either direction

I'm defining my circle parametrically counterclockwise

[queen herald]

Yes, it will change the sign

[queen herald]

Ok then (y', -x') points outwards

[quaint field]

does flux have a sign?

I feel like it's a scalar

so ig it wouldn't matter the direction?

[queen herald]

Basically it counts how much volume flows in and out per unit of time

If there's more "fluid" flowing out than flowing in, then the flux is positive, and if more fluid flows in than out, then the flux is negative

[quaint field]

ohh

so I guess you want to define your normal as pointing "out"

[queen herald]

yes

[quaint field]

how do you know how to manipulate the velocity vector to produce a normal pointing out?

do you just look at the graph?

or I guess it depends on clockwise/counterclockwise as you mentioned

[queen herald]

Honestly I just multiply the velocity vector by $\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$ (which corresponds to a clockwise rotation by $\pi/2$) and normalise the result

[grand pond]

EQUENOS

[queen herald]

Yes

It indeed does

[quaint field]

so if you're moving CCW then you need a 90 degree CW rotation and vice versa?

[queen herald]

yes

[quaint field]

ok nice

wait, the normal vector of the circle would be the same direction as the radius vector right

that makes it easy to find then

[queen herald]

If the circle is centered at the origin then yes

[quaint field]

but even if not, you can consider the radius vector from the center of the circle right

[queen herald]

Oh yeah that's correct

I thought you meant like (x(t), y(t))

[quaint field]

I need to plug in r(t) into my flux as well right

because my flux is in terms of x and y

[queen herald]

What's r(t)?

I thought r in the task is some constant

[quaint field]

I assumed r = x^2 + y^2

but r(t) is just my position vector on the circle

nothing to do with r

am I doing this correctly? I'm getting a complicated integral ,lol

also I guess I am using R instead of a

@queen herald I'm getting some complicated integral

[queen herald]

I think we don't need v(t) here

It's just the integral of F•n

So the flux should be just $$\int_0^{2\pi} \frac{R + \cos t}{1+R^2+2R\cos t} dt$$

[grand pond]

EQUENOS

[queen herald]

Anyways, it's still a similar integral

,w integrate (R + cos(t))/(1+R^2+2 R cos(t)) dt

[grand pond]

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[quaint field]

but isn't ds = v dt

also is there a simpler form for that definite integral somehow

[queen herald]

Oh sorry I forgot about ds

Yes you're right

[quaint field]

idk if this helps

https://math.stackexchange.com/questions/134577/integrate-int-frac1a-cos-x-dx

[queen herald]

Idk, but after substitutions it should look relatively simple

[quaint field]

also how would it be different when radius is less than 1

I feel like I would be doing the same thing

[queen herald]

Geometrically the difference is very clear:
When radius is more than 1, the vector field always points outwards from the circle
If the radius is less than 1, the field points inwards on the left part of the circle and outwards on the right part

[quaint field]

yes

[queen herald]

It's the same formula though

[quaint field]

but I won't have to modify the actual procedure right?
it'll take care of itself with the sign change